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3 years ago

What number is a common factor for 16 and 56

Mathematics

2 answers:

Daniel [21]3 years ago

6 0

8 is the common factor for these numbers

valina [46]3 years ago

3 0

8 is the greatest common factor

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Sugarcrash lyrics u will get brainliest (u can look it up if u want)

gulaghasi [49]

I’m on a sugar crash I ain’t got no f cash

Maybe I should take a bath cut my f brain in half

I’m not lonely just a bit tired of this f sh

Nothing that I write can make me

Feel good

Victim of the great machine in love with everything I see

Neon lights surrounding me I indulge in luxury

Everything I do is wrong ‘cept for when I hit the bong

(Hit the bong, hit the b-b-b-b-b)

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Feeling sh in my bed didn’t take my f meds

Hype pop up in my ears everything just disappears

Don’t wanna be someone else just don’t wanna hate myself

I just don’t wanna hate myself instead I wanna

Feel good

I’m on a sugar crash I ain’t got no f cash

Maybe I should take a bath cut my f brain in half

I’m not lonely just a bit tired of this f sh

Nothing that I write can make me

Feel good

Where to now?

Got the rest of my life just spread out

Got the rest of my life to f around

Got the rest of my life to make sound

5 0

2 years ago

Read 2 more answers

The highest point in colorado is mount elbert, at 14,433 feet.About how many miles is that

Juli2301 [7.4K]

2.7335227 because you know how ,any ft. are in 1 mile so you just multiply :)

8 0

3 years ago

I need help with both of them I'm confused...

irga5000 [103]

The first question is a and the second is a

5 0

3 years ago

Read 2 more answers

Find the surface area of the solid generated by revolving the region bounded by the graphs of y = x2, y = 0, x = 0, and x = 2 ab

Nikitich [7]

Answer:

See explanation

Step-by-step explanation:

The surface area of the solid generated by revolving the region bounded by the graphs can be calculated using formula

$SA=2\pi \int\limits^a_b f(x)\sqrt{1+f'^2(x)} \, dx$

If $f(x)=x^2,$ then

$f'(x)=2x$

and

$b=0\\ \\a=2$

Therefore,

$SA=2\pi \int\limits^2_0 x^2\sqrt{1+(2x)^2} \, dx=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx$

Apply substitution

$x=\dfrac{1}{2}\tan u\\ \\dx=\dfrac{1}{2}\cdot \dfrac{1}{\cos ^2 u}du$

Then

$SA=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx=2\pi \int\limits^{\arctan(4)}_0 \dfrac{1}{4}\tan^2u\sqrt{1+\tan^2u} \, \dfrac{1}{2}\dfrac{1}{\cos^2u}du=\\ \\=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0 \tan^2u\sec^3udu=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0(\sec^3u+\sec^5u)du$

Now

$\int\limits^{\arctan(4)}_0 \sec^3udu=2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17})\\ \\ \int\limits^{\arctan(4)}_0 \sec^5udu=\dfrac{1}{8}(-(2\sqrt{17}+\dfrac{1}{2}\ln(4+\sqrt{17})))+17\sqrt{17}+\dfrac{3}{4}(2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17}))$

Hence,

$SA=\pi \dfrac{-\ln(4+\sqrt{17})+132\sqrt{17}}{32}$

3 0

3 years ago

3b 2a=5a−6 solve the equation for a in terms of <br> b.

oee [108]

Is it 3b-2a=5a-6
or
3b+2a=5a-6

you are missing the sign of 2a

4 0

3 years ago